Question: Simplify the following expression: $y = \dfrac{-3x^2- 7x- 4}{-3x - 4}$
Explanation: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(-3)}{(-4)} &=& 12 \\ {a} + {b} &=& &=& {-7} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $12$ and add them together. The factors that add up to ${-7}$ will be your ${a}$ and ${b}$ When ${a}$ is ${-4}$ and ${b}$ is ${-3}$ $ \begin{eqnarray} {ab} &=& ({-4})({-3}) &=& 12 \\ {a} + {b} &=& {-4} + {-3} &=& -7 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({-3}x^2 {-4}x) + ({-3}x {-4}) $ Factor out the common factors: $ x(-3x - 4) + 1(-3x - 4)$ Now factor out $(-3x - 4)$ $ (-3x - 4)(x + 1)$ The original expression can therefore be written: $ \dfrac{(-3x - 4)(x + 1)}{-3x - 4}$ We are dividing by $-3x - 4$ , so $-3x - 4 \neq 0$ Therefore, $x \neq -\frac{4}{3}$ This leaves us with $x + 1; x \neq -\frac{4}{3}$.